Question: Multiply the following rational expressions and simplify the result. $\dfrac{k^{12}+19k^6+88}{5k^8+40k^2} \cdot \dfrac{k^4-6k^2}{2k^{12}+19k^6-33}=$
Solution: Let's first factor the numerators and denominators of each expression separately. [Why are we doing this?] The numerator, $k^{12}+19k^6+88$, of the first expression can be factored as $(k^6+8)(k^6+11)$ using the sum-product pattern. The denominator, $5k^8+40k^2$, of the first expression can be factored as $5k^2(k^6+8)$ by factoring out $5k^2$. The numerator, $k^4-6k^2$, of the second expression can be factored as $k^2(k^2-6)$ by factoring out $k^2$. The denominator of the second expression $2k^{12}+19k^6-33$ can be factored by grouping to $(2k^6-3)(k^6+11)$. Now the product looks as follows: $\dfrac{(k^6+8)(k^6+11)}{5k^2(k^6+8)}\cdot\dfrac{k^2(k^2-6)}{(2k^6-3)(k^6+11)}$ To find the product of two rational expressions, we multiply across, then simplify: [What's that?] $\phantom{=}\dfrac{(k^6+8)(k^6+11)}{5k^2(k^6+8)}\cdot\dfrac{k^2(k^2-6)}{(2k^6-3)(k^6+11)}$ $\begin{aligned} &=\dfrac{(k^6+8)(k^6+11) \cdot k^2(k^2-6)}{5k^2(k^6+8) \cdot (2k^6-3)(k^6+11)} &\text{Multiply across.}\\\\\\ &= \dfrac{{\cancel{(k^6+8)}}{\cancel{(k^6+11)}} {\cancel{k^2}}(k^2-6)}{5{\cancel{k^2}}{\cancel{(k^6+8)}} (2k^6-3){\cancel{(k^6+11)}}} &\text{Cancel out common factors.}\\\\\\\\ &=\dfrac{k^2-6}{5(2k^6-3)} \end{aligned}$ Therefore, the simplified form of the product is $\dfrac{k^2-6}{5(2k^6-3)}$, which is equivalent to $\dfrac{k^2-6}{10k^6-15}$.